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READING QUIZ

1 The angular of a rotating two-dimensional rigid CEE 271: Applied II, body about its center of G is . (a) mvG – Lecture 33: Ch.19, Sec.1–3 – (b) IGvG (c) mω (d) IGω Prof. Albert S. Kim ANS: (d) 2 If a rotates about a fixed axis passing through its Civil and Environmental , University of Hawaii at Manoa center of mass, the body’s linear momentum is . (a) a constant Date: ______(b) zero (c) mvG (d) IGω ANS: (b)

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LINEAR AND , APPLICATIONS PRINCIPLE OF AND MOMENTUM

Today’s objectives: Students In-class activities: • The swing bridge opens and will be able to • Reading Quiz closes by turning using a 1 Develop formulations for • Applications motor located under the the linear and angular • Linear and Angular center of the deck at A that momentum of a body. Momentum applies a M to the 2 Apply the principle of linear • Principle of Impulse and bridge. and angular impulse and Momentum momentum. • Concept Quiz • If the bridge was supported at its end B, would the same • Group Problem Solving torque open the bridge in the same , or would it open • Attention Quiz slower or faster? • What are the benefits of making the bridge with the variable depth (thickness) substructure as shown?

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PROCEDURE FOR ANALYSIS EXAMPLE (Solution)

• Impulse-momentum diagram: Mt W t

1 Establish the x, y, z inertial . mvG1 mvG2 + = 2 Draw the impulse-momentum diagrams for the body. F t 3 I ω I ω Compute IG, as necessary. G 1 Nt G 2 4 Apply the equations of impulse and momentum (one vector and one scalar or the three scalar equations). • Impulse and Momentum using : (vG)2 = rω2: t2 5 If more than three unknowns are involved, kinematic (HA)1 + Σ MAdt = (HA)2 equations relating the of the mass center G and Zt1 the ω should be used to furnish additional 0 + Mt = m(vG)2r + IGω2 2 2 equations. = mr ω2 + m(kO) ω2 2 2 = m r + (kO) ω2  300(6)  ω2 (t = 6 s) = = 11.5 rad/s 300(0.62 + 0.42) 13 / 36 15 / 36

EXAMPLE CONCEPT QUIZ

• Given: The 300 kg wheel has a 1 If a slab is rotating about its center of mass radius of gyration about its mass G, its angular momentum about any center O of kO = 0.4 meter. The arbitrary point P is its angular wheel is subjected to a momentum computed about G (i.e., IGω). · of M = 300 N m. (a) larger than (b) less than (c) the same as • Find: The angular velocity after 6 seconds if it starts from (d) None of the above rest and no slipping occurs. ANS: (c) • Plan: Time as a parameter should make you think Impulse 2 The linear momentum of the slab in question 1 is . and Momentum! Since the body rolls without slipping, point (a) constant A is the center of rotation. Therefore, applying the angular (b) zero impulse and momentum relationships along with (c) increasing linearly with time kinematics should solve the problem. (d) decreasing linearly with time ANS: (b) 14 / 36 16 / 36

ATTENTION QUIZ CONSERVATION OF MOMENTUM

Today’s objectives: Students In-class activities: will be able to • Reading Quiz 1. If a slender bar rotates about end A, its angular 1 Understand the conditions • Applications momentum with respect to A is? for conservation of linear • Conservation of Linear and and angular momentum. Angular Momentum 1 A (a) ml2ω 2 Use the condition of  12 • Concept Quiz  (b) 1 ml2ω conservation of linear/  6 • Group Problem Solving l 1 2 angular momentum.  (c) 3 ml ω • Attention Quiz G  (d) ml2ω ω  ANS: (c)  

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ATTENTION QUIZ (Continued) READING QUIZ

1 If there are no external impulses acting on a body . (a) only linear momentum is conserved (b) only angular momentum is conserved 2. As in the principle of and , if a does no (c) both linear momentum and angular momentum are work, it does not need to be shown on the impulse and conserved momentum diagram/equation. (d) neither linear momentum nor angular momentum are (a) False conserved (b) True ANS: (c) (c) Depends on the case 2 If a rigid body rotates about a fixed axis passing through its (d) No clue! center of mass, the body’s linear momentum is . ANS: (a) (a) a constant (b) zero (c) mvG (d) IGω ANS: (b)

22 / 36 24 / 36 APPLICATIONS CONSERVATION OF LINEAR MOMENTUM (19.3)

• Recall that the linear impulse and momentum relationship is t2 L1 + F dt = L2 Z X t1 where L1 = (mvG)1 and L2 = (mvG)2. • If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms • A skater spends a lot of time either spinning on the ice or are zero. Thus, the linear momentum for a rigid body (or rotating through the air. To spin fast, or for a long time, the system) is constant, or conserved. So L1 = L2. skater must develop a large amount of angular momentum. • This equation is referred to as the • If the skater’s angular momentum is constant, can the skater conservation of linear momentum. The conservation of vary her rotational ? How? linear momentum equation can be used if the linear • The skater spins faster when the arms are drawn in and impulses are small or non-impulsive. slower when the arms are extended. Why?

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APPLICATIONS(continued) CONSERVATION OF ANGULAR MOMENTUM

• Similarly, if the sum of all the angular impulses due to external acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved

t2 HG1 + MGdt = HG2 Z X t1

where HG1 = IGω1 and HG2 = IGω2. • The resulting equation is referred to as the conservation of • Conservation of angular momentum allows cats to land on angular momentum or HG1 HG2. their feet and divers to flip, twist, spiral and turn. It also = helps teachers make their heads spin! • If the initial condition of the rigid body (or system) is known, conservation of momentum is often used to determine the • Conservation of angular momentum makes final linear or angular velocity of a body just after an event water circle the drain faster as it gets closer to the drain. occurs.

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GROUP PROBLEM SOLVING ATTENTION QUIZ

• Given: Two children 1. Using conservation of linear and angular momentum (mA = mB = 30 kg) sit at the requires that edge of the merry-go-round, (a) all linear impulses sum to zero which has a mass of 180 kg and (b) all angular impulses sum to zero (c) both linear and angular impulses sum to zero a radius of gyration of kz = 0.6 m. (d) None of the above ANS: (c) 2. The angular momentum of a body about a point A that is the fixed axis of rotation but not the mass center (G) is • Find: The angular velocity of the merry-go-round if A jumps (a) IAω off horizontally in the +t direction with a speed of 2 m/s, (b) IGω measured relative to the merry-go-round. (c) rG(mvG) + IGω • Plan: Draw an impulse-momentum diagram. The (d) Both (a) and (c) conservation of angular momentum can be used to find the ANS: (d) angular velocity.

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GROUP PROBLEM SOLVING(Solution) Note

• Apply the conservation of angular momentum equation: M B A H1 = H2 2 X X 180(0.6) (2) + 2 × [(30)2(0.75)2] = 180(0.6)2ω + (30)ω(0.75)2 +(30)(0.75ω + 2)(0.75) vA/M A M Now we solve B 197.1 = 98.55ω + 45 ω = 1.54 rad/s

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